In $\textit{Set Theory}$ by Jech 1978 edition, in the proof of Lemma 32.5 which you can hopefully see at the google book link

http://books.google.com/books?id=pLxq0myANiEC&pg=PA389&lpg=PA389&dq=weakly+compact+less+than+measurable&source=bl&ots=0DjaJGx_1D&sig=Av3IvVVUAkmQgUly0jNqbGN5Pz4&hl=en&sa=X&ei=fcADU-L4JIyBogTwwoGAAg&ved=0CCwQ6AEwAA#v=onepage&q=weakly%20compact%20less%20than%20measurable&f=false

In the course of the proof using the tree property, he produces from any weakly compact cardinal $\kappa$ a non principal $L_\alpha$-ultrafilter $U$ on $\kappa$, which is $L_\alpha$-$\kappa$ complete, and moreover the intersection of $\kappa$ many elements of $U$ (taken in $V$) is nonempty. 

The latter implies that countable intersection (taken in $V$) is nonempty. This fact implies iterability for example by 19.13 of Kanamori. So $L_\alpha$ with this ultrafilter can be iterated of length $\text{Ord}$

However, $0^\sharp$ follows from the existence of a mouse (iterable premouse). But $0^\sharp$ has stronger consistency strength than a weakly compact.

Is this $(L_\alpha, U)$ not a pre-mouse? I have found varying definition of pre-mouse. One definition has the addition condition that $\kappa$ should be the largest cardinal of $L_\alpha$. This seems like it may not hold since at the beginning of the proof, the proof chooses an $\alpha$ such that $L_\alpha \models ZF^-$. 

I would like to have the $\text{Ord}$ length iteration for what I am trying to do; however, I am troubled by whether or not this would give an iterable premouse and hence imply sharps. 


Thanks for any clarification you can provide.